If z=x+iy and real part `((z-1)/(2z+i))=1` then locus of z is

Options

A. 1. circle whose diameter is `(sqrt(5))/(2)`
B. 2. straight line whose slope is `3/2`
C. 3. straight line whose slope is `-2/3`
D. 4. circle whose centre is at `(-(1)/(2), -(3)/(2))` .

Answer

To solve the problem, we need to find the locus of the complex number \( z = x + iy \) given that the real part of the expression \( \frac{z - 1}{2z + i} = 1 \).

### Step-by-Step Solution:

1. **Substituting \( z \)**:
\[ z = x + iy \implies z - 1 = (x - 1) + iy \]
\[ 2z + i = 2(x + iy) + i = 2x + (2y + 1)i \]

2. **Setting up the equation**:
We need to find the real part of:
\[ \frac{(x - 1) + iy}{2x + (2y + 1)i} \]
We set this equal to 1:
\[ \text{Re}\left(\frac{(x - 1) + iy}{2x + (2y + 1)i}\right) = 1 \]

3. **Rationalizing the denominator**:
Multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{(x - 1) + iy}{2x + (2y + 1)i} \cdot \frac{2x - (2y + 1)i}{2x - (2y + 1)i} \]
The denominator becomes:
\[ (2x)^2 + (2y + 1)^2 = 4x^2 + (2y + 1)^2 \]
The numerator becomes:
\[ (x - 1)(2x) - (y(2y + 1)) + i[(x - 1)(-(2y + 1)) + (y)(2x)] \]

4. **Expanding the numerator**:
The real part of the numerator is:
\[ 2x(x - 1) - y(2y + 1) \]
The imaginary part is:
\[ (y)(2x) - (x - 1)(2y + 1) \]

5. **Setting the real part equal to 1**:
Now we have:
\[ \frac{2x(x - 1) - y(2y + 1)}{4x^2 + (2y + 1)^2} = 1 \]
Cross-multiplying gives:
\[ 2x(x - 1) - y(2y + 1) = 4x^2 + (2y + 1)^2 \]

6. **Rearranging the equation**:
Rearranging leads to:
\[ 2x^2 - 2x - 2y^2 - y - 4x^2 - 4y - 1 = 0 \]
Simplifying gives:
\[ -2x^2 - 2y^2 + 2x + 3y + 1 = 0 \]
Dividing through by -2:
\[ x^2 + y^2 - x - \frac{3}{2}y - \frac{1}{2} = 0 \]

7. **Completing the square**:
Completing the square for \( x \) and \( y \):
\[ (x - \frac{1}{2})^2 + (y - \frac{3}{4})^2 = \frac{5}{4} \]

8. **Identifying the locus**:
This represents a circle centered at \( \left(\frac{1}{2}, \frac{3}{4}\right) \) with a radius of \( \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \).

### Final Result:
The locus of \( z \) is a circle whose diameter is \( \sqrt{5}/2 \).